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3: \(P=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(z+y\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{y+x}\right)+\dfrac{1}{4}\left(\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)=\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = x = \(\dfrac{1}{3}\).
Ta có (x+y)xy=x2+y2-xy
=> \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\)
<=>\(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2+\frac{3}{4}\left(\frac{1}{x}-\frac{1}{y}\right)^2\ge\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
<=> \(0\le\frac{1}{x}+\frac{1}{y}\le4\)
mà \(A=\frac{1}{x^3+y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)
Vậy Max A =16 khi \(x=y=\frac{1}{2}\)
Theo đề ta suy ra \(y\le1-3x\)
\(\Rightarrow\sqrt{xy}\le\sqrt{x\left(1-3x\right)}\)
Ta có \(A=\frac{1}{x}+\frac{1}{\sqrt{xy}}\ge\frac{1}{x}+\frac{1}{\sqrt{x\left(1-3x\right)}}\ge\frac{1}{x}+\frac{1}{\frac{x+\left(1-3x\right)}{2}}=\frac{2}{2x}+\frac{2}{-2x+1}\)
\(=2\left(\frac{1}{2x}+\frac{1}{-2x+1}\right)\ge2.\frac{\left(1+1\right)^2}{2x-2x+1}=8\)
Vậy \(A\ge8\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}x=1-3x=y\\\frac{1}{2x}=\frac{1}{-2x+1}\\3x+y=1\end{cases}}\) \(\Leftrightarrow\) \(x=y=\frac{1}{4}\)
Từ giả thiết ta có: \(x+y+z=xyz\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Ta có:
\(M=\frac{\left(x-1\right)+\left(y-1\right)}{y^2}-\frac{1}{y}+\frac{\left(y-1\right)+\left(z-1\right)}{z^2}-\frac{1}{z}+\frac{\left(z-1\right)+\left(x-1\right)}{x^2}-\frac{1}{x}\)
\(=\left[\frac{\left(x-1\right)}{y^2}+\frac{\left(x-1\right)}{x^2}\right]+\left[\frac{y-1}{y^2}+\frac{y-1}{z^2}\right]+\left[\frac{z-1}{z^2}+\frac{z-1}{x^2}\right]-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\left(x-1\right)\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\left(y-1\right)\left(\frac{1}{y^2}+\frac{1}{z^2}\right)+\left(z-1\right)\left(\frac{1}{z^2}+\frac{1}{x^2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\ge\frac{2\left(x-1\right)}{xy}+\frac{2\left(y-1\right)}{yz}+\frac{2\left(z-1\right)}{zx}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2\)
Lại có:
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\ge3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=3\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\sqrt{3}\)
\(\Rightarrow M\ge\sqrt{3}-2\)
Dấu bằng xảy ra khi x=y=z=\(\sqrt{3}\)
Từ điều kiện suy ra \(\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)
Áp dụng BĐT Cô-si, ta có :
\(3\le\sqrt{xy}+\sqrt{x}.1+\sqrt{y}.1\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}\)
\(\Rightarrow x+y\ge2\)
Ta có : \(\frac{x^2}{y}+y\ge2\sqrt{\frac{x^2}{y}.y}=2x\); \(\frac{y^2}{x}+x\ge2\sqrt{\frac{y^2}{x}.x}=2y\)
\(\Rightarrow\frac{x^2}{y}+\frac{y^2}{x}+x+y\ge2x+2y\)
\(\Rightarrow P=\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\ge2\)
Vậy GTNN của P là 2 khi x = y = 1